Quant - Number System Ex. 2

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#1 vasu on 01.09.09 at 4:30 pm

preparation material

#2 vasu on 01.09.09 at 4:33 pm

In Preparation Material
Exercise2 + Answers + Explanations

Question:
‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse?

(1) 13
(2) 23
(3) 37
(4) 41

Correct Answer - (2)

Explanation:
The value of the square of the hypotenuse = h2 = a2 + b2

As the problem states that ‘a’ and ‘b’ are positive integers, the values of a2 and b2 will have to be perfect squares. Hence we need to find out that value amongst the four answer choices which cannot be expressed as the sum of two perfect squares.

Choice 1 is 13. 13 = 9 + 4 = 32 + 22. Therefore, Choice 1 is not the answer as it is a possible value of h2

Choice 2 is 23. 23 cannot be expressed as the sum two numbers, each of which in turn happen to be perfect squares. Therefore, Choice 2 is the answer.

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Question:
Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

(1) 599
(2) 1021
(3) 263
(4) Cannot be determined

Correct Answer - (1)

Explanation:
Two numbers when divided by a common divisor, if they leave remainders of x and y and when their sum is divided by the same divisor leaves a remainder of z, the divisor is given by x + y - z.

In this case, x and y are 431 and 379 and z = 211. Hence the divisor is 431 + 379 - 211 = 599.

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Question:
What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?

(1) 144
(2) 72
(3) 108
(4) 216

Correct Answer - (2)

Explanation:
100! has 348 as the greatest power of 3 that can divide it. Similarly, the greatest power of 2 that can divide 100! is 297. 297 = 448 * 21.
Therefore, the largest power of 12 that can divide 100! is 48.
Therefore, for 350 to be included in 100!, 100! needs to be multiplied by 32 * 23 = 9 * 8 = 72.

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Question:
A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11, viz 88?

(1) 3
(2) 21
(3) 59
(4) 68

Correct Answer - (3)

Explanation:
When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7.
Therefore, d1r2 + r1 = 8*7 + 3 = 59.

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Question:
What is the total number of different divisors including 1 and the number that can divide the number 6400?

(1) 24
(2) 27
(3) 27
(4) 68

Correct Answer - (2)

Explanation:
To find the number of divisors for a number, express then number as the product of the prime numbers like ax * by * cz .

In this case, 6400 can be expressed as 64*100 = 26 * 4 * 25 = 28 * 52.

Having done that, the way to find the number of divisors is by multiplying the indices of each of the prime numbers after incrementing the indices by 1.
i.e. the number of divisors = (x+1)(y+1)(z+1).
In this case, (8 + 1)(2 + 1) = 9*3 = 27.

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#3 prashant on 03.12.09 at 3:30 pm

VERY VERY GOOD

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